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Let $C([0,1])$ be the space of all continuous function on $[0,1]$. Define the an inner product in the space: $$ \langle f,g \rangle = \int_0^{1}f(t)g(t)\;dt $$ Let $K$ be a subspace of $C$: $$ K=\{ax+bx^2;a,b \in \mathbb{R}\} \subset C $$ How do I find the orthogonal complement subspace of K? More generally, what is a general method to find an orthogonal complement of a subspace?

Chinnapparaj R
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A Slow Learner
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  • It's difficult to express any simpler than ${x}^\perp \cap {x^2}^\perp$. It will be infinite-dimensional, after all. – Theo Bendit Mar 24 '19 at 07:49
  • @TheoBendit so we cannot write an explicit form for functions in this subspace? – A Slow Learner Mar 24 '19 at 09:18
  • No, not as far as I know. What would an explicit form look like anyway, given that writing it as a span is out of the question? – Theo Bendit Mar 24 '19 at 09:49
  • @TheoBendit This is not part of the question, but the reason I am asking this question is because I am looking for a function in $K$ that is closet to the constant function 1, with respect to the norm defined by the inner product. Could you help me find that function? – A Slow Learner Mar 24 '19 at 19:43
  • Yes, I absolutely can. If you project the constant function $1$ onto $K$, then $1$ minus this projection will be the projection onto $K^\perp$. You should find an orthonormal basis for $K$, by applying Gram-Schmidt to your basis $(x, x^2)$. If $(e_1, e_2)$ is your orthonormal basis, the projection onto $K$ will be $\langle 1, e_1 \rangle e_1 + \langle 1, e_2 \rangle e_2$. – Theo Bendit Mar 24 '19 at 19:56
  • @TheoBendit so the function in $K$ closet to the constant function 1 will be its projection onto $K$, not the projection onto its orthogonal complement (1-the other), right? If so, why did you mention the orthogonal complement in your last comment? – A Slow Learner Mar 25 '19 at 01:28
  • I'm saying that $P_K(1) + P_{K^\perp}(1) = 1$. If you can find the projection $P_K(1)$ of $1$ onto $K$, then you'll get the projection onto $K^\perp$ basically for free. If you want me to write a full answer, I suggest asking a new question. – Theo Bendit Mar 25 '19 at 01:33

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